解
an=1-n²吗?
sn=a1+a2+a3+……+an
=(1-1²)+(1-2²)+(1-3²)+……+(1-n²)
=n-(1²+2²+3²+……+n²)
=n-n(n+1)(2n+1)/6
=[6n-n(n+1)(2n+1)]/6
这是要记住1²+2²+3²+……+n²=n(n+1)(2n+1)/6